Radical, Log and Exponent Inequalities - Problems
1. (i) Graph y1 = and y2 = x (ii) Solve: < x
2. Solve:
3. Solve:
4. Solve:
5. Solve: (Hint: y = ax (a > 1) is an increasing function)
6. Solve: log4(x + 7) > log2(x + 1)
7. Solve:
8. Solve:
9. Prove: If p ≥ 1, then 5p2 − 1 ≥ 4p
10.
11. 12. logx(x + 6) > 2logxx
13. Graph x2 − 3xy + 2y2 < 0 14. Graph
15. Graph (x − y2)(x − 4y2) > 0 16.
17. 18.
19. 20.
21. 22.
23. 24.
25. 26.
27. x < x2 − 12 < 4x 28.
29.
Radical, Log and Exponent Inequalities – Solutions
1. (i) graph
2. breaks down into two cases:
(i) or (ii)
The plan is to 'exponentiate' both sides (keeping in mind 'log base 3' is increasing).
or
Check out the graph of y = | log3|x| |
y = 1
3.
Case (i) 7 − x > 0 (x < 7) Case (ii) x ≥ 7
Here we can square both sides Here, 7 − x ≥ 0, so the statement is always true.
x + 5 > 49 − 14x + x2
0 > x2 − 15x + 44 Hence x ≥ 7 are all solutions.
0 > (x − 11)(x − 4)
4 < x < 11 and x < 7 gives…
solutions: 4 < x < 7
Taking the union of (i) and (ii) we get:
4. − 1 ≤ x ≤ 3
5. Now we'll note y = 2x is an increasing function…
Now case (i) x + 1 > 0 ( x > −1)
−2x2 − 2x > 2x (no inequality-sign switch!)
−2x2 − 4x > 0 2x2 + 4x < 0 2x(x + 2) < 0
−2 < x < 0 but recall x > −1 so…
[Note: x + 1 0] The intersection gives us: − 1 < x < 0
case (ii) x + 1 < 0 ( x < −1)
Just switch the inequality sign to get 2x(x + 2) > 0
From (i) and (ii) x > 0 or x < −2 combined with x < −1
gives us: x < −2
Radical, Log and Exponent Inequalities – Solutions
6. log4(x + 7) > log2(x + 1)
Now before squaring both sides… case (i) x + 1 > 0 (x > −1)
x + 7 > x2 + 2x + 1 0 > x2 + x − 6
0 > (x + 3)(x − 2) gives…
−3 < x < 2 and recall x > −1
Taking the intersection we get: −1 < x < 2
Before getting to case (ii) x + 1 ≤ 0 , note the restrictions on the arguments for log functions, viz., x + 1 > 0 and x + 7 > 0, hence case (ii) has no solutions. We're done.
7. x2 + 1 < 2 (sign switch!)
Since 'log base ½ ' is a decreasing function: y1 > y2 iff x1 < x2 , hence the sign switch!
Continuing: x2 − 1 < 0 iff (x−1)(x+1) < 0 iff −1 < x < 1
8. Restrictions: x + 1 > 0 and because '1+x' is
case (i) increasing (1+x > 1) also a 'base', it can't be zero or one, so
which is to say: x > 0, then we have also we have 2 − x > 0 ( x < 2)
2 − x < 1 + x
1 < 2x case (ii) 0 < 1 + x < 1 (decreasing function)
x > ½ and x < 2 (restriction) or −1 < x < 0 and switch the '<' sign to get:
½ < x < 2 x < ½ and −1 < x < 0 intersecting…
(i) and (ii) give: − 1 < x < 0
9. p2 − 4p − 1 ≥ 0 (5p + 1)(p − 1) ≥ 0 x ≤ −1/5 or x ≥ 1
Hence if p ≥ 1, then 5p2 − 1 ≥ 4p (Notice the converse is not true.)
10. Multiply both sides by x2 > 0 (x) to get:
x4 − 1 > x3 − x x4 − x3 + x − 1 > 0 x3(x − 1) + (x − 1) > 0
(x3 + 1)(x − 1) > 0 Now use our number line trick to get: x < −1 or x > 1
11. 1 < x < 3
12. logx(x + 6) > 2logxx Restrictions: x > 0, x and x + 6 > 0 (x > −6)
Case (i) logx increasing iff x > 1 , so x + 6 > x2 0 > x2 − x − 6 = (x + 2)(x − 3)
−2 < x < 3 and x > 1 give us: 1 < x < 3
Case (ii) logx decreasing iff 0 < x < 1 and then (switching the inequality sign)…
x > 3 or x < −2 and 0 < x < 1 give us: empty set!
13. Factor x2 − 3xy + 2y2 < 0 to get:
(x − y)(x − 2y) < 0
Case (i) x − y > 0 and x − 2y <0
y < x and y > ½ x (1st Quadrant)
Case (ii) x − y <0 and x − 2y >0
y > x and y < ½ x (3rd Quadrant)
Radical, Log and Exponent Inequalities – Solutions
14. (Inner radius 1, outer radius 2)
15. (x − y2)(x − 4y2) > 0
Case (i) + + y2 < x and 4y2 < x Shade left of both sideways parabolas (see above) and
take the intersection.
Case (ii) − − y2 > 0 and 4y2 > x Shade right of both parabolas (narrow area above).
16. = Restrictions: x > 1
'log base 3/10' is a decreasing function, hence: x − 1 >
Squaring both sides x2 − 2x + 1 > x − 1
(x − 2)(x − 1) = x2 − 3x + 2 > 0
Number line… x > 2 or x < 1 but recall x > 1 , so…
x > 2
17. Let's get the variable on top.
18. | x | ≤1/2 ,
19.
20. Hence, −9/2 ≤ x < −3
21.
getting rid of the '−' sign…
22.
23. | 5 − 3x | ≤ 20 or squaring 25 − 30x + 9x2 ≤ 400
−20 ≤ 5 − 3x ≤ 20 9x2 − 30x − 375 ≤ 0
−25 ≤ −3x ≤ 15 (9x − 75)(x + 5) ≤ 0
−15 ≤ 3x ≤ 25 −5 ≤ x ≤ 25/3
−5 ≤x ≤ 25/3
Radical, Log and Exponent Inequalities – Solutions
24. Squaring both sides: x2 + 10x + 25 < x2 −2x + 1
12x + 24 < 0 iff x < −2
25. Square both sides, then clear of fractions by multiplying by x2.
25 − 10/x + 1/x2 < 1
25x2 − 10x + 1 < x2 iff 24x2 − 10x + 1 < 0 iff (6x − 1)(4x − 1) < 0
1/6 < x < ¼
26. Square both sides.
x4 − 4x2 + 4 ≤ 1 iff x4 − 4x2 + 3 = (x2 − 3)(x2 − 1) ≤ 0
iff iff
27. x < x2 − 12 and x2 − 12 < 4x
0 < x2 − x − 12 x2 − 4x − 12 < 0
0 < (x + 3)(x − 4) (x + 2)(x − 6) < 0
x < −3 or x > 4 −2 < x < 6
Intersecting these two cases, we get: 4 < x < 6
28.
x2 − 4x + 4 < 4(x2 + 6x + 9) Careful not to forget to square the '2' also.
0 < 3x2 + 28x + 32 = (3x + 4)(x + 8)
x < −8 , x > −4/3
29.
Case (i) x ≥ 0 Case (ii) x < 0
(since x < 0, 2x − 1 < 0 also)
(Now let u = 2x )
x ≥ 0 and gives 1 + 2u(u) − 2u ≤ 0 (Clearing of fractions)
But 2u2 − 2u + 1 is never zero (always positive)
since is not defined.
Only the first case yields: No solutions in this case.